I'm talking about a neutral hydrogen atom here:

  1. According to Bohr, if the electron jumps from $n=2$ to $n=1$, there'd be only one wavelength of light being emitted. However, that is not the case we see experimentally. When an electron jumps from $n=1$ to $n=2$, more than one frequency of light, the values of which will be very close to each other (say $500nm$ and $500.01nm$) can be emitted. This is because, in the same shell, there are subshells, and some subshells have more energy than others. So, if the electron jumps from the $2s$ subshell to the $1s$ subshell, the emitted light's wavelength will be say $500.01nm$, and if the electron jumps from $2p$ to $1s$ the emitted light's wavelength will be say $500nm$. This is because the energy difference between $1s$ and $2s$ and $1s$ and $2p$ aren't the same. So, when an electron jumps from $n=2$ to $n=1$, there won't be a single line in the emission spectra as predicted by Bohr, rather there will be multiple fine lines very close to each other.
  2. Bohr's model can't explain the splitting of spectral lines into multiple fine lines under the influence of an electric or magnetic field. The spectral lines split because within subshells, there are orbitals, and they are differently oriented three-dimensionally. This has significant electromagnetic implications. For example, a $p$-subshell has three orbitals $p_x,p_y$ & $p_z$. They are differently oriented three-dimensionally, and as they have electrons moving in them, they have different electromagnetic orientations as well even though their energies are the same: they are degenerate orbitals. So, normally, in an emission spectrum, you don't notice any difference between $p_x,p_y$ & $p_z$ as they have the same energy, but under the influence of an electric or a magnetic field, you start to appreciate their individuality, as the lines emitted split. I think when observing the fine line of a $p$-subshell, if we apply an electric or a magnetic field, the line of the $p$-subshell will split into three finer lines with the same wavelength.


  1. Is my explanation as a whole correct?

PS: In reality, there are no such things as shells or subshells. There are only orbitals. To make the learning curve less steep, teachers use these terms to high-school students like us.


2 Answers 2


Sentence n.1

Comparison of any theory with experimental values of spectral lines width is always a non-trivial task. This statement is true for Bohr's theory and Quantum Mechanics in the non-relativistic or relativistic formulation or even the full QED for the isolated H atom. There is the possibility, and actually, this is the case, that some important effect is not included. All the isolated atom mechanisms of broadening of the spectral lines of the Hydrogen atom in the absence of an external field are usually smaller than Doppler's and pressure broadening. The former is due to the doppler shift of the line frequency due to the spread of thermal velocities. The latter, also known as collision broadening, is due to the interaction between atoms when they get closer. Notice that both the effects would be present even at the level of Bohr's theory and would give a broadening of spectral lines larger than one order of magnitude in the majority of the experimental conditions.

The mechanism you are considering in your statement 1. is what one would call the intrinsic linewidth of the Hydrogen atom spectrum. In a complete QM treatment, its origin is due, as you correctly state, to the partial removal of the non-relativistic energy level degeneracy of the two opposite charges problem. Essential effects are due to the spin-orbit coupling (fine structure), electron spin-proton spin interaction (hyperfine structure), and QED effects (Lamb shift). However, this statement, too, would need a necessary correction. Indeed, removal of degeneracy would result in sharp lines with a small separation. Instead, even ignoring the spurious (but unavoidable) broadening due to the experimental finite resolution, there is always an intrinsic linewidth due to the finite time required by every electronic transition. According to the time-energy Uncertainty Principle, a transition occurring in a time $\tau$ would correspond to an intrinsic broadening $\Delta E$ not smaller than $\frac{\hbar}{\tau}$.

Sentence n.2

Shells and subshells are not equivalent to orbitals. They are just old names for energy levels. Similarly, the word orbital is just a different name for eigenfunction. Every orbital has corresponding energy, although one energy level may correspond to different orbitals. This is the degeneracy of the energy levels.

Within Bohr's theory there are no orbitals but quantized orbits. Apart from this warning, one can speak about energy levels and their degeneracy as in modern Quantum Mechanics. Actually, after Sommerfeld extended Bohr's ideas to elliptic orbits, the degeneracy of the energy levels was the correct one. The only problem is connected to the presence of the spin and its consequences. I think that in your statement 2. the presence of the spin should be mentioned explicitly.

  • $\begingroup$ Thanks for your awesome reply; upvoted! I edited my question slightly: the ordering; thought I'd let you know. $\endgroup$ Nov 3, 2021 at 11:14

First of all, I just want to point out that the orbitals typically taught in high school are actually not quite convenient to work with when trying to understand the hydrogen spectra. Rather than thinking in terms of the "real orbitals" $p_x$, $p_y$, and $p_z$, it's actually more correct to think about it in terms of the orbitals with particular a magnetic quantum number $m_l = -1, 0, 1$. In my answer, I'm going to ignore this distinction because I don't think it adds much value to be precise about here, but see this question on real vs complex orbitals if you're curious.

  1. It is true that there are multiple fine spectral lines if you look at the transitions from $n=2$ to $n=1$, but not because the $2s$ and $2p$ orbitals have different energies per se.

    For a hydrogen atom, the orbitals for the same $n$ have identical energy, e.g. $2s$ and $2p$ orbitals have the same energy if we ignore other effects I mention below. This is one of the classic "accidental degeneracies", which is named as such because it's not obvious it should be the case. When you have more than one electron in the atom (so any other neutral atom), the degeneracy in the orbital energies is broken because of the electron-electron interactions, and what you are saying would be true.

    So what does cause the splitting of the spectral lines? There are actually multiple effects at play:
  • First, there is fine structure arising from spin orbit coupling (interaction between the electron spin and the orbital angular momentum) as well as some relativistic corrections.
  • If you look even closer, there is Lamb shift that comes from properly treating the electric field quantum mechanically.
  • If you look really closely, you will see hyperfine structure that arises from interaction between the proton spin and the electron spin. Even the ground state $1s$ orbital is affected and splits into two different energies.
  1. The application of electric or magnetic fields does break degeneracies, although how it works is a bit different for electric vs magnetic.

    The Stark effect is the splitting of spectral lines under the application of an electric field. For hydrogen, the Stark effect does result in 3 lines for $n=2$, but it's not exactly true that it's one line per p-subshell. In some sense, two of the p-orbitals actually have the same energy, and then the s-orbital, which for hydrogen has the same energy as a p-orbital, mixes with one of the p-orbitals to create two "hybrid" orbitals that have two different energies. There's a much longer and proper description in this question.

    The Zeeman effect is the splitting of spectral lines under the application of a magnetic field. Roughly speaking, this will split the energies of the p-orbitals in the way you are imagining, but depending on the strength of the magnetic field, it can also split the two possible electron spin states. If we ignore fine structure, this means that at high magnetic fields, you actually get five lines from the $2s$ and $2p$ orbitals, not four, which you might otherwise expect.
  • $\begingroup$ Awesome answer, thanks! $\endgroup$ Nov 3, 2021 at 11:48

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